Hi! I was trying to do this last exercise of 2016 exam, but I can't understand the solution to the 1st question.
In particular, given the derivative wrt to \(b_{u'}\) that is reported in the solution, I would have written the optimal \(b_{u'}\) as: \(b_{u'}=\frac{\sum_{u'~m} r_{u'm}-v_{u'}^T w_m - b_{u'}-b_m}{\lambda}\)
But in the solution, at the denominator there is also \(\sum_{u'~m} 1\) and I don't understand where this element come from.
Thanks in advance for you answer!
deriving \( 1/2 \Sigma (f_{um} -r_{um})^2 \) gives you \( 1/2 \Sigma 2(<v_u, w_m> + b_u +b_m - r_{um}) \) by chain rule, and deriving the second part of the objective gives you only \(\frac{\lambda}{2} 2 b_u \). Factorize: from the first part you can split the terms of the sum into separate sums, one for \( b_u \) and one the rest of the terms, and you have \( b_u \) as a constant within the corresponding sum so it makes the sum(1) appear.
Matrix factorization exercise
Hi! I was trying to do this last exercise of 2016 exam, but I can't understand the solution to the 1st question.
In particular, given the derivative wrt to \(b_{u'}\) that is reported in the solution, I would have written the optimal \(b_{u'}\) as:
\(b_{u'}=\frac{\sum_{u'~m} r_{u'm}-v_{u'}^T w_m - b_{u'}-b_m}{\lambda}\)
But in the solution, at the denominator there is also \(\sum_{u'~m} 1\) and I don't understand where this element come from.
Thanks in advance for you answer!
1
deriving \( 1/2 \Sigma (f_{um} -r_{um})^2 \) gives you \( 1/2 \Sigma 2(<v_u, w_m> + b_u +b_m - r_{um}) \) by chain rule, and deriving the second part of the objective gives you only \(\frac{\lambda}{2} 2 b_u \). Factorize: from the first part you can split the terms of the sum into separate sums, one for \( b_u \) and one the rest of the terms, and you have \( b_u \) as a constant within the corresponding sum so it makes the sum(1) appear.
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